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Sergio039 [100]
3 years ago
11

How many moles of ScCl3 can be produced when 10.00 mol Sc react with 9.00 mol Cl2

Chemistry
2 answers:
WITCHER [35]3 years ago
8 0

Answer:

6.00 moles of ScCl3 will be produced.

Explanation:

Step 1: Data given

Moles of Sc = 10.00 moles

Moles of Cl2 = 9.00 moles

Molar mass of Sc = 44.96 g/mol

Molar mass of Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Sc + 3Cl2 → 2ScCl3

Step 3: Calculate the limiting reactant

For 2 moles Sc we need 3 moles Cl2 to produce 2 moles ScCl3

Cl2 is the limiting reactant. It will completely be consumed (9.00 moles)

Sc is the limiting reactant. There will react 2/3 * 9.00 = 6.00 moles

There will remain 10.00 - 6.00 =4.00 moles Sc

Step 4: Calculate moles ScCl3

For 3 moles Cl2 we'll have 2 moles ScCl3

For 9.00 moles we'll have 6.00 moles of ScCl3

6.00 moles of ScCl3 will be produced.

Nuetrik [128]3 years ago
4 0

Answer:

Moles of ScCl_3 = 6 moles

Explanation:

The reaction of Sc and Cl_2 to make ScCl_3 is:

2Sc+3Cl_2⇒2ScCl_3

The above reaction shows that 2 moles of Sc  can react with 3 moles of Cl_2 to form ScCl_3.

Mole Ratio= 2:3

For 10 moles of Sc we need:

Moles of Cl_2 = Moles of Sc *\frac{3 moles of Cl_2}{2 Moles of Sc}

Moles of Cl_2 = 10 *\frac{3 moles of Cl_2}{2 Moles of Sc}

Moles of Cl_2 =15 moles

So 15 moles of Cl_2 are required to react with 10 moles of Sc but we have 9 moles of Cl_2 , it means Cl_2 is limiting reactant.

Moles of ScCl_3=Given\  Moles\  of\ Cl_2 *\frac{2\  Moles\ o\ fScCl_3}{3\ Moles\ of\ Cl_2}

Moles\ of\  ScCl_3=9 *\frac{2\  Moles\ of\ ScCl_3}{3\ Moles\ of\ Cl_2}

Moles of ScCl_3= 6 moles

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